One diagonal of a rhombus is decreasing at a rate of $7$ centimeters per minute and the other diagonal of the rhombus is increasing at a rate of $10$ centimeters per minute. At a certain instant, the decreasing diagonal is $4$ centimeters and the increasing diagonal is $6$ centimeters. What is the rate of change of the area of the rhombus at that instant (in square centimeters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $16$ (Choice C) C $-1$ (Choice D) D $-16$ The area of a rhombus with diagonals $d_1$ and $d_2$ is $\dfrac{d_1d_2}{2}$.
Explanation: Setting up the math Let... $d_1(t)$ denote the decreasing diagonal of the rhombus at time $t$, $d_2(t)$ denote the increasing diagonal of the rhombus at time $t$, and $A(t)$ denote the area of the rhombus at time $t$. We are given that $d_1'(t)=-7$ and $d_2'(t)=10$ (notice that $d_1'$ is negative). We are also given that that $d_1(t_0)=4$ and $d_2(t_0)=6$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures The measures relate to each other through the formula for the area of a rhombus: $A(t)=\dfrac{d_1(t)d_2(t)}{2}$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=\dfrac{d_1'(t)d_2(t)+d_1(t)d_2'(t)}{2}$ Using the information to solve Let's plug ${d_1'(t_0)}={-7}$, ${d_2(t_0)}={6}$, ${d_1(t_0)}={4}$, and $C{d_2'(t_0)}=C{10}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=\dfrac{{d_1'(t_0)}{d_2(t_0)}+{d_1(t_0)}C{d_2'(t_0)}}{2} \\\\ &=\dfrac{({-7})({6})+({4})(C{10})}{2} \\\\ &=-1 \end{aligned}$ In conclusion, the rate of change of the area of the rhombus at that instant is $-1$ square centimeters per minute. Since the rate of change is negative, we know that the area is decreasing.